Difference between revisions of "2025 AMC 8 Problems/Problem 12"

Latest revision as of 22:25, 31 January 2025

Problem

The region shown below consists of 24 squares, each with side length 1 centimeter. What is the area, in square centimeters, of the largest circle that can fit inside the region, possibly touching the boundaries?

$[asy] size(100); void drawSquare(pair p) { draw(box(p, p + (1,1)), black); } int[][] grid = { {0, 0, 0, 0, 0, 0}, {0, 0, 1, 1, 0, 0}, {0, 1, 1, 1, 1, 0}, {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, {0, 1, 1, 1, 1, 0}, {0, 0, 1, 1, 0, 0}, {0, 0, 0, 0, 0, 0} }; int rows = grid.length; int cols = grid[0].length; for (int i = 0; i < rows; ++i) { for (int j = 0; j < cols; ++j) { if (grid[i][j] == 1) { drawSquare((j, rows - i - 1)); } } } [/asy]$

$\textbf{(A)}\ 3\pi\qquad \textbf{(B)}\ 4\pi\qquad \textbf{(C)}\ 5\pi\qquad \textbf{(D)}\ 6\pi\qquad \textbf{(E)}\ 8\pi$

Solution 1

The largest circle that can fit inside the figure has its center in the middle of the figure and will be tangent to the figure in $8$ points. By the Pythagorean Theorem, the distance from the center to one of these $8$ points is $\sqrt{2^2 + 1^2} = \sqrt5$ , so the area of this circle is $\pi \sqrt{5}^2 = \boxed{\textbf{(C)} 5\pi}$ .

~Soupboy0

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/VP7g-s8akMY?si=Oe8Ka0kEZiLNXA5g&t=1096 ~hsnacademy

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI

Video Solution 2 by Thinking Feet

https://youtu.be/PKMpTS6b988

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-==Problem==
+== Problem ==
 The region shown below consists of 24 squares, each with side length 1 centimeter. What is the area, in square centimeters, of the largest circle that can fit inside the region, possibly touching the boundaries?
 <asy>
-import graph;
 size(100);
-pen gridPen = black;
 void drawSquare(pair p) {
-     draw(box(p, p + (1,1)), gridPen);
+     draw(box(p, p + (1,1)), black);
 }
@@ Line 38: / Line 35: @@
 <math>\textbf{(A)}\ 3\pi\qquad \textbf{(B)}\ 4\pi\qquad \textbf{(C)}\ 5\pi\qquad \textbf{(D)}\ 6\pi\qquad \textbf{(E)}\ 8\pi</math>
+== Solution 1 ==
+The largest circle that can fit inside the figure has its center in the middle of the figure and will be tangent to the figure in <math>8</math> points. By the Pythagorean Theorem, the distance from the center to one of these <math>8</math> points is <math>\sqrt{2^2 + 1^2} = \sqrt5</math>, so the area of this circle is <math>\pi \sqrt{5}^2 = \boxed{\textbf{(C)} 5\pi}</math>.
-==Solution==
+~Soupboy0
-The largest circle that can fit inside the figure has its center in the middle of the figure and will be tangent to the figure in <math>8</math> points. The distance from the center to one of these <math>8</math> points can be found with the Pythagorean Theorem: <math>r^2 = 2^2 + 1^2 \rightarrow r = \sqrt5</math>. Therefore, the area of this circle = <math>\pi (\sqrt{5^2}) = \boxed{\textbf{(C)} 5\pi}</math>.
+==Video Solution (A Clever Explanation You’ll Get Instantly)==
+https://youtu.be/VP7g-s8akMY?si=Oe8Ka0kEZiLNXA5g&t=1096
+~hsnacademy
-~Soupboy0
+== Video Solution 1 by SpreadTheMathLove ==
-==Video Solution 1 by SpreadTheMathLove==
 https://www.youtube.com/watch?v=jTTcscvcQmI
-==Video Solution by Thinking Feet==
+== Video Solution 2 by Thinking Feet ==
 https://youtu.be/PKMpTS6b988
-==See Also==
+== See Also ==
 {{AMC8 box|year=2025|num-b=11|num-a=13}}
 {{MAA Notice}}
+[[Category:Introductory Geometry Problems]]

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